Percentage Questions with Solutions — Set 4 (Q31 to Q40)

This set contains Percentage Questions with Solutions — Set 4 (Q31 to Q40) covering a mix of question types and difficulty levels — from basic to advanced — exactly as asked in real competitive exams.

Solutions are written in a simple, step-by-step notebook style for easy self-study and quick understanding. Each solution is broken down step by step so even the toughest question feels easy. These questions are hand-picked for students preparing for SSC CGL, SSC CHSL, CAT, Bank PO, Bank Clerk, UPSC CSAT, Railway RRB, AMCAT, eLitmus, TCS NQT and all campus placement aptitude tests. International students preparing for GRE, GMAT, SAT, ACT, MAT and all Numerical Reasoning Tests will find these equally useful.

✏️ Attempt each question on your own first — then check the solution below.

31. Two numbers x & y are such that the sum of 6% of x and 5% of y is \(\frac{3}{5}\) of the sum of 6% of x and 7% of y. Then find the ratio of x and y.

32. The ratio of number of boys to that of girls in a school is 5 : 1. If 70% of boys and 60% of girls are scholarship holders, then find the percentage of students who do not get scholarship.

33. The cost of manufacture of an article is made up of 4 components A, B, C and D which have a ratio of 1 : 2 : 3 : 4 respectively. If there are respective changes in the ost of 10%, -20%, -30%, -40%, then what will be the percentage change in the cost ?

34. The height of a triangle is increased by 20%. What could be maximum percentage increase in length of base whose area increases by 60%.

35. In an examination there are three subjects physics, chemistry and math having max marks 200, 120, 150 respectively. A student gets 50%, 70%, 60% in physics, chemistry, math respectively. If he wants to get 70% marks if four subjects then how many marks he must obtain in biology of max marks 200.

36. In an examination 500 girls and 1000 boys appeared. In which 50% girls and 40% boys passed the exam. Then find the % of failed students ?

37. In a class, the number of girls is 20% more than that of the boys. The strength of the class is 77. If 6 more girls are admitted to the class then find the ratio of the number of boys to that of the girls.

38. If 70% of the students in a school are girls and the number of boys is 570, how many girls are in school ?

39. Ankit invests 1000 Rs. in some share in the ratio 2 : 3 : 5 which pay dividends of 5%, 10% and 20% (on his investment) for that year respectively. Find the dividend income.

40. A batsman scored 130 runs which included 5 four and 5 sixes. What percentage of his total score did he make by running between the wickets ?

Check your answers below. Each solution is written exactly the way a student writes in their own notebook — clear, simple and easy to follow.

31. Two numbers x & y are such that the sum of 6% of x and 5% of y is \(\frac{3}{5}\) of the sum of 6% of x and 7% of y. Then find the ratio of x and y.
Sol:
6% × x + 5% × y = \(\frac{3}{5}\)(6% × x + 7% × y)
⟹ \(\frac{6}{{100}}x + \frac{5}{{100}}y = \frac{3}{5}(\frac{6}{{100}}x + \frac{7}{{100}}y)\)
⟹ 30x + 25y = 18x + 21y
⟹ 12x = – 4y
3x = -y
∴ \( \boldsymbol {\frac{x}{y} = – \frac{1}{3}} \)     Answer

32. The ratio of number of boys to that of girls in a school is 5 : 1. If 70% of boys and 60% of girls are scholarship holders, then find the percentage of students who do not get scholarship.
Sol:
\(\frac{{boys}}{{girls}} = \frac{5}{1} = \frac{{500}}{{100}}\)
∴ total students who do not get scholarship
 = 500(100 – 70)% + 100(100 – 60)%
= 500 × \(\frac{{30}}{{100}}\) + 100 × \(\frac{{40}}{{100}}\)
= 150 + 40
= 190
∴ required number of percentage of students who do not get scholarship = \(\frac{{190}}{{600}} \times 100\) = \( \boldsymbol{\frac{{19}}{6}} \)%      Answer

33. The cost of manufacture of an article is made up of 4 components A, B, C and D which have a ratio of 1 : 2 : 3 : 4 respectively. If there are respective changes in the cost of 10%, -20%, -30%, -40%, then what will be the percentage change in the cost ?
Sol:

total old cost = 100 + 200 + 300 + 400
                        = 1000
total new cost = 110 + 160 + 390 + 240
                          = 900
∴ required % change = \(\frac{{900 – 1000}}{{1000}} \times 100\% \)
= – 10%      Answer

34. The height of a triangle is increased by 20%. What could be maximum percentage increase in length of base whose area increases by 60%.
Sol:
\(60 = x + 20 + \frac{{20x}}{{100}}\)
\(40 = x + \frac{x}{5}\)
200 = 6x
⟹ x = \(\frac{{200}}{6} = \frac{{100}}{3}\) = \( \boldsymbol{33\frac{1}{3}} \)%        Answer

35. In an examination there are three subjects physics, chemistry and math having max marks 200, 120, 150 respectively. A student gets 50%, 70%, 60% in physics, chemistry, math respectively. If he wants to get 70% marks if four subjects then how many marks he must obtain in biology of max marks 200.
Sol:

Total max marks of 4 subject = 200 + 120 + 150 + 200
= 670
∴ 70% of max marks = \(\frac{{70}}{{100}} \times 670\)
= 469
& total marks obtained in three subjects
= 200 × 50% + 120 × 70% + 150 × 60%
= 100 + 84 + 90
= 274

∴ marks to be obtained in biology = 469 – 274 = 195        Answer

36. In an examination 500 girls and 1000 boys appeared. In which 50% girls and 40% boys passed the exam. Then find the % of failed students ?
Sol:

∴ % of failed students = \(\frac{{250 + 600}}{{500 + 1000}} \times 100\)
= \(\frac{{850}}{{1500}} \times 100\)
= \(\frac{{17}}{3}\)
= \( \boldsymbol{5\frac{2}{3}} \)%     Answer  

37. In a class, the number of girls is 20% more than that of the boys. The strength of the class is 77. If 6 more girls are admitted to the class then find the ratio of the number of boys to that of the girls.
Sol:

Strength of class = 5 + 6 = 11 ⟶ 77 (Given)
∴ Number of girls = 6 × 7 = 42
& Number of boys = 5 × 7 = 35
∴ \(\frac{{boys}}{{girls}} = \frac{{35}}{{42}}\) = \( \boldsymbol{\frac{5}{6}} \)      Answer

38. If 70% of the students in a school are girls and the number of boys is 570, how many girls are in school ?
Sol:

39. Ankit invests 1000 Rs. in some share in the ratio 2 : 3 : 5 which pay dividends of 5%, 10% and 20% (on his investment) for that year respectively. Find the dividend income.
Sol:
Ratio of shares = 2x : 3x : 5x
∴ 2x + 3x + 5x = 1000
x = 100
∴ 1ˢᵗ share = 200
2ⁿᵈ share = 300
3ʳᵈ share = 500
& dividend income = \(200 \times \frac{5}{{100}} + 300 \times \frac{{10}}{{100}} + 500 \times \frac{{20}}{{100}}\)
 = 10 + 30 + 100
 =  140         Answer

40. A batsman scored 130 runs which included 5 four and 5 sixes. What percentage of his total score did he make by running between the wickets ?
Sol:

The runs scored buy running = 130 – 5 × 4 – 5 × 6 = 80
∴ required % = \(\frac{{80}}{{130}} \times 100\)
= \( \boldsymbol{61\frac{7}{{13}}} \)%       Answer

✅ Well done on completing Set 4!

Continue practising with Percentage Questions 41 to 50 → Set 5 or revisit the Percentage Concept Page to strengthen your formulas and tricks before moving ahead.

Consistent practice is the key to mastering percentage for SSC CGL, SSC CHSL, CAT, Bank PO, Bank Clerk, UPSC CSAT, Railway RRB, AMCAT, eLitmus, TCS NQT and international exams including GRE, GMAT, SAT, ACT, MAT and all Numerical Reasoning Tests. Want to understand the concept better? Read about Percentage on Wikipedia before attempting the next set.

This page is part of our complete series of percentage questions with solutions for competitive exams — covering every question type from basic to advanced so you can build speed, accuracy and confidence. Practising these questions regularly will also strengthen your core percentage concept before your exam day.